Displacement, distance, velocity, speed and acceleration in one dimension, with a sign convention

The five constant-acceleration formulae, including s = vt - ½at², and how to choose between them

Vertical motion under gravity with g = 9.8 m s⁻²

Journeys made of several stages, each with its own constant acceleration

Displacement–time, velocity–time, speed–time and acceleration–time graphs: gradients and areas

Trapezium journeys, unknown times found from areas, and two-vehicle (overtaking and meeting) problems

1. Straight-line kinematics models a moving body as a particle with constant acceleration and gives two tools for it, the five suvat formulae and the velocity–time graph; the hardest questions need both.

2. Choose a positive direction and give every displacement, velocity and acceleration a sign.

3. With constant acceleration, list s, u, v, a, t, mark the three you know, and pick the formula without the one you neither know nor want.

4. Under gravity the acceleration is 9.8 m s⁻² downwards all the way: on the way up, at the top and on the way down.

5. Multi-stage journeys: the final velocity of one stage is the initial velocity of the next.

6. Gradient of s–t = velocity; gradient of v–t = acceleration; area under v–t = displacement; area under a–t = change in velocity. Two objects meet when their displacements from the same origin are equal, not when their velocities are.

In M1 the moving body is modelled as a particle: its size is ignored, so it has a single position on a line. Positions are measured from a fixed origin O, and one direction along the line is chosen as positive.

A deceleration (or retardation) of 2 m s⁻² means an acceleration of 2 m s⁻² against the motion: write a = -2 if the motion is in the positive direction.

Average velocity = total displacement ÷ total time; average speed = total distance ÷ total time. A round trip has zero average velocity but a non-zero average speed.

“At rest” and “instantaneously at rest” both mean v = 0. A particle can be instantaneously at rest while it still has an acceleration, at the top of a throw for example.

The constant-acceleration (suvat) formulae

s = displacement (m); u = initial velocity (m s⁻¹); v = final velocity (m s⁻¹); a = acceleration (m s⁻²); t = time (s). Not in the formulae list: all five must be known. Valid only while the acceleration is constant; if it changes, split the motion into stages.

Each formula leaves out exactly one of the five quantities. Write down s, u, v, a, t, fill in the three you know and the one you want, and use the formula that leaves out the fifth.

Where they come from. Constant acceleration means the velocity–time graph is a straight line with gradient a: (v - u)/t = a, which is v = u + at. The area under it is a trapezium: s = ½(u + v)t. Substituting v = u + at into the area gives s = ut + ½at²; substituting u = v - at gives s = vt - ½at²; eliminating t gives v² = u² + 2as.

WORKED EXAMPLE · QUESTION

A cyclist moving along a straight road at 3 m s⁻¹ accelerates uniformly at 0.75 m s⁻² for 8 s. (a) Find her speed at the end of the 8 s. (b) Find the distance she travels in the 8 s. (c) Find the time she takes to travel the first 20 m.

SOLUTION

(a) u = 3, a = 0.75, t = 8, want v: v = 3 + 0.75 ×8 = 9 m s⁻¹

(b) s = ½(3 + 9) ×8 = 48 m

(c) s = 20, u = 3, a = 0.75, want t: the formula without v. 20 = 3t + 0.375t² ⇒0.375t² + 3t - 20 = 0, so t = (-3 + √(9 + 30))/0.75 = 4.33 s (3 s.f.). The other root is negative and is rejected: the motion starts at t = 0.

When a quadratic gives two positive roots. If u and a have opposite signs, both roots of s = ut + ½at² can be positive: the particle passes the point, stops, comes back and passes it again. Both times are valid if the acceleration really stays the same for that long; read the question to decide.

WORKED EXAMPLE · QUESTION

A particle P moves along a straight line with constant acceleration. It passes a point A and, 5 s later, passes a point B, where AB = 45 m. At B its speed is 4 m s⁻¹, in the direction from A to B. (a) Find the acceleration of P. (b) Find the speed of P at A. (c) The acceleration stays the same after P passes B. Find the times after passing A at which P is 24 m from A, and interpret them.

SOLUTION

(a) s = 45, v = 4, t = 5, want a; u is unknown, so use the formula without u: 45 = 4 ×5 - ½a ×25 ⇒12.5a = -25 ⇒ a = -2 m s⁻² (a deceleration of 2 m s⁻²)

(b) 4 = u - 2 ×5 ⇒ u = 14 m s⁻¹. Check with the formula without a: ½(14 + 4) ×5 = 45 ✓.

(c) From A, u = 14, a = -2, s = 24: 24 = 14t - t² ⇒t² - 14t + 24 = 0 ⇒(t - 2)(t - 12) = 0. t = 2 s and t = 12 s. At t = 2, v = 14 - 4 = +10: moving away from A. At t = 12, v = 14 - 24 = -10: P has stopped (at t = 7, 49 m from A) and is on its way back.

A body moving freely up or down near the Earth's surface is modelled as a particle with no air resistance, so its acceleration is constant: g = 9.8 m s⁻² vertically downwards. This is true on the way up, at the highest point and on the way down. At the top, v = 0 but a = -9.8 (up positive).

Take up as positive for anything thrown upwards: then a = -9.8, and a point below the start has a negative displacement. For something only falling, down positive (a = +9.8) is fine.

For a particle projected upwards at u: time to the top = u/g; greatest height above the start = u²/2g (from v² = u² + 2as with v = 0).

Symmetry: it returns to its starting level after 2u/g, with the same speed u it was projected with.

Using g = 9.8, give answers to 2 or 3 significant figures.

WORKED EXAMPLE · QUESTION

A stone is thrown vertically upwards at 14.7 m s⁻¹ from a point on a cliff 44.1 m above the sea. It falls past the cliff into the sea. The stone is modelled as a particle moving freely under gravity. (a) Find the greatest height of the stone above the sea. (b) Find the time from throwing until the stone reaches the sea. (c) Find its speed as it reaches the sea. (d) Find the length of time for which the stone is above the point of projection.

SOLUTION

(a) Up positive. At the top v = 0; u = 14.7, a = -9.8: 0 = 14.7² - 2 ×9.8 ×s ⇒s = 11.025 m. Height above the sea = 44.1 + 11.025 = 55.1 m

(b) The sea is 44.1 m below the start: s = -44.1. -44.1 = 14.7t - 4.9t² ⇒t² - 3t - 9 = 0 ⇒t = (3 + √45)/2 = 4.85 s

(c) Using v² = u² + 2as, which avoids the rounded time: v² = 14.7² + 2 ×(-9.8) ×(-44.1), so the speed is 32.9 m s⁻¹

(d) Back at the point of projection (s = 0) when 14.7t - 4.9t² = 0: t = 14.7 ÷4.9 = 3 s (= 2u/g)

Split the motion wherever the acceleration changes: an engine cuts out, brakes are applied, a string breaks, the particle hits the ground. Work through the stages in order. The final velocity of one stage is the initial velocity of the next, and the displacements add. Keep the same positive direction and origin for every stage. A sketch of the velocity–time graph is the best plan: one straight segment per stage.

WORKED EXAMPLE · QUESTION

A model rocket is launched vertically upwards from rest at ground level. While its motor burns, it moves with constant acceleration 6 m s⁻² upwards. The motor stops after 4 s, and the rocket then moves freely under gravity until it hits the ground. (a) Find the speed and height of the rocket when the motor stops. (b) Find the greatest height reached. (c) Find the total time from launch until the rocket hits the ground.

SOLUTION

(a) Stage 1: u = 0, a = 6, t = 4. v = 6 ×4 = 24 m s⁻¹; s = ½ ×6 ×4² = 48 m. 24 m s⁻¹ at 48 m

(b) Stage 2 starts with u = 24 (up positive), a = -9.8; at the top v = 0: 0 = 24² - 19.6s ⇒s = 29.39 m. Greatest height = 48 + 29.39 = 77.4 m

(c) Stage 2 from the motor stopping to the ground: s = -48. -48 = 24t - 4.9t² ⇒4.9t² - 24t - 48 = 0 ⇒t = 6.423 s (positive root). Total time = 4 + 6.423 = 10.4 s

Check by another route: 2.449 s up to the top, then 3.974 s falling 77.39 m from rest: 4 + 2.449 + 3.974 = 10.42 s ✓. Stage 2 is one constant-acceleration stage from the motor stopping all the way to the ground; there is no need to split it at the top.

On an s–t graph a straight line means constant velocity, a horizontal line means at rest, and a curve means the velocity is changing. A maximum or minimum of s is where v = 0.

On a v–t graph a straight segment means constant acceleration. Where the graph crosses the t-axis, the particle changes direction.

Displacement = area above the axis − area below; distance = area above + area below. A speed–time graph is the v–t graph with the parts below the axis reflected above it.

For constant-acceleration stages the a–t graph is a set of horizontal segments; each rectangle's area is the change in velocity in that stage.

WORKED EXAMPLE · QUESTION

A particle moves along a straight line, starting from rest at O. Its velocity–time graph consists of straight segments joining (0, 0), (4, 8), (10, 8), (16, −4) and (20, −4), with t in seconds and v in m s⁻¹. (a) Find the acceleration in each of the four stages. (b) Find the time at which the particle is instantaneously at rest. (c) Find the displacement from O at t = 20. (d) Find the total distance travelled in the 20 s.

SOLUTION

(a) Gradients: 8 ÷4 = 2 m s⁻²; 0; (-4 - 8) ÷6 = -2 m s⁻²; 0. 2, 0, −2, 0 m s⁻²

(b) The third segment crosses v = 0: from 8 at t = 10, losing 2 m s⁻¹ each second, t = 10 + 8 ÷2 = 14 s

(c) Above: ½ ×4 ×8 + 6 ×8 + ½ ×4 ×8 = 16 + 48 + 16 = 80 m. Below: ½ ×2 ×4 + 4 ×4 = 4 + 16 = 20 m. Displacement = 80 - 20 = 60 m

(d) Distance counts both areas as positive: 80 + 20 = 100 m

Many journeys are “accelerate – steady speed – decelerate”: a trapezium on the v–t graph. Label every time and speed on a sketch, using letters (T, V) for the unknowns. Then write gradient equations for the accelerations (time = change in speed ÷ acceleration) and an area equation for the distance. Area of a trapezium = ½ × (sum of parallel sides) × height; for the whole journey, ½ × (time at steady speed + total time) × top speed. If the unknown appears in a product, the area equation may be a quadratic: check which root fits the times on your sketch.

WORKED EXAMPLE · QUESTION

A train starts from rest at station P and accelerates uniformly at 0.5 m s⁻² until its speed is 20 m s⁻¹. It then travels at 20 m s⁻¹ for T seconds, and then decelerates uniformly at 0.8 m s⁻², coming to rest at station Q. The distance PQ is 4.2 km. (a) Sketch the velocity–time graph. (b) Find T. (c) Find the average speed for the whole journey. (d) Find the time after leaving P at which the train has travelled 1 km.

SOLUTION

(a) A trapezium: up from (0, 0) to 20 m s⁻¹, horizontal for T s, down to zero; the deceleration line is steeper than the acceleration line (see the figure).

(b) Accelerating: 20 ÷0.5 = 40 s; decelerating: 20 ÷0.8 = 25 s. Area: ½ ×40 ×20 + 20T + ½ ×25 ×20 = 4200 ⇒650 + 20T = 4200 ⇒ T = 177.5 s

(c) Total time = 40 + 177.5 + 25 = 242.5 s; average speed = 4200 ÷242.5 = 17.3 m s⁻¹

(d) 400 m are covered in the first 40 s, so 1 km is reached at steady speed: 400 + 20(t - 40) = 1000 ⇒ t = 70 s

Measure both displacements from the same origin and the same starting time. If one object starts later, its time is (t − delay).

They are level (one overtakes the other, or they meet) when their displacements are equal. Coming towards each other from D apart, they meet when the distances they have each travelled add up to D.

The gap between them is greatest (or least) when their velocities are equal, where the v–t lines cross.

If one object changes stage, check which stage it is in at the meeting time: a root that lies outside the stage you used is not valid. On a v–t graph the meeting time is when the areas under the two graphs are equal.

WORKED EXAMPLE · QUESTION

A car is at rest at traffic lights. At t = 0, a lorry passes the car moving at a constant 15 m s⁻¹, and at the same instant the car moves off in the same direction with constant acceleration 2.5 m s⁻² until it reaches 25 m s⁻¹. It then continues at 25 m s⁻¹. (a) Show that the car has not caught up with the lorry when it reaches its top speed. (b) Find the time at which the car overtakes the lorry, and the distance from the traffic lights at that moment. (c) Find the greatest distance between the lorry and the car before the car overtakes.

SOLUTION

(a) The car reaches 25 m s⁻¹ after 25 ÷2.5 = 10 s, having travelled ½ ×2.5 ×10² = 125 m. The lorry after 10 s: 15 ×10 = 150 m >125 m, so the lorry is still ahead.

(b) For t ≥10, car: 125 + 25(t - 10); lorry: 15t. 125 + 25t - 250 = 15t ⇒10t = 125 ⇒ t = 12.5 s, at 187.5 m from the lights

(c) The gap grows while the lorry is faster and shrinks once the car is faster, so it is greatest when the speeds are equal: 2.5t = 15 ⇒t = 6. Gap = 15 ×6 - ½ ×2.5 ×6² = 90 - 45 = 45 m

Solving ½ ×2.5t² = 15t gives t = 12 s, but that assumes the car is still accelerating after 10 s, which it is not. Part (a) is there to stop you using that root.

P1 Pure Mathematics 1

Finding times from s = ut + ½at² is solving a quadratic; reading gradients and areas uses straight-line graphs and the area of a trapezium.

M1.3 Dynamics

A constant resultant force gives a constant acceleration (F = ma), which is what makes the suvat formulae usable for connected particles, slopes and particles that slide and then collide.

M1.1 Vectors, M2 and IAL Physics

Motion in a plane at constant velocity is written as r = r₀ + vt; M2 uses calculus when the acceleration varies and treats projectiles as two straight-line motions at once. IAL Physics Unit 1 uses the same equations, usually with g = 9.81 m s⁻².

1. A small stone is dropped from rest and falls 2.5 m to the floor. Find its speed as it reaches the floor.

2. A car decelerates uniformly at 1.5 m s⁻² and comes to rest after 12 s. Find the distance it travels in those 12 s, without first finding its initial speed.

3. A velocity–time graph is a straight line from (0, 10) to (5, 0). State the acceleration and the displacement over the 5 s.

4. A ball is projected vertically upwards from ground level at 9.8 m s⁻¹. Find how long it is in the air, and explain why its mass is not needed.

5. A particle has velocity v = 6 - 2t m s⁻¹ for 0 ≤t ≤5. Find its displacement and the distance it travels in the 5 s.

1. Down positive: v² = 0 + 2 ×9.8 ×2.5 = 49, so v = 7 m s⁻¹ (exactly, with g = 9.8).

2. s = vt - ½at² with v = 0, a = -1.5, t = 12: s = 0 + ½ ×1.5 ×144 = 108 m.

3. Gradient = (0 - 10) ÷5 = −2 m s⁻² (a deceleration of 2 m s⁻²). Area = ½ ×5 ×10 = 25 m.

4. 0 = 9.8t - 4.9t² ⇒t = 2 s. Without air resistance every body falls with the same acceleration g, so the mass does not appear in the model.

5. It stops at t = 3 after 9 m; from t = 3 to 5 it moves back ½ ×2 ×4 = 4 m. Displacement = 9 - 4 = 5 m (also 6 ×5 - 25); distance = 9 + 4 = 13 m.